A particle moves along the curve $y=-2x^4+10$ so that that the $x$ -coordinate is increasing at a constant rate of $\dfrac12$ units per second. What is the magnitude (in units per second) of the particle's velocity vector when the particle is at the point $(-1,8)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sqrt{65}}{2}$ (Choice B) B $\dfrac{79}{8}$ (Choice C) C $\dfrac{\sqrt{257}}{4}$ (Choice D) D $\dfrac{\sqrt{257}}{2}$
Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This however shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dx}{dt}=\dfrac12$ for any value of $t$. We are asked for the magnitude of the particle's velocity vector when the particle is at the point $(-1,8)$. In other words, we are asked for $\left|\left|\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)\right|\right|$ at the point $(-1,8)$. Finding $\dfrac{dy}{dt}$ $\dfrac{dy}{dt}=-4x^3$ Finding $\dfrac{dy}{dt}$ at $(-1,8)$ The expression for $\dfrac{dy}{dt}$ only depends on the particle's $x$ -coordinate, which in our case is ${x}={-1}$ : $\begin{aligned} \dfrac{dy}{dt}&=-4({-1})^3 \\\\ &=4 \end{aligned}$ Therefore, the particle's velocity vector at the point $(-1,8)$ is $\left(\dfrac12,4\right)$. Finding $\left|\left|\left(\dfrac12,4\right)\right|\right|$ $\left|\left|\left(\dfrac12,4\right)\right|\right|=\dfrac{\sqrt{65}}{2}$ In conclusion, the magnitude of the particle's velocity vector when the particle is at the point $(-1,8)$ is $\dfrac{\sqrt{65}}{2}$ units per second.